Problem Description

In this problem, we are given a string s. Our task is to make the smallest palindrome by adding characters at the front of s. A palindrome is a string that reads the same forward and backward. The key here is that we can only add characters to the beginning of s to form the palindrome; we cannot modify or add characters at the end. We need to find and return the shortest possible palindrome that can be obtained from s through such transformations.

Intuition

To solve this problem effectively, we need to identify the longest palindrome that starts at the beginning of s. Once we find such a palindrome, we can mirror the remaining part of the string (that isn't included in the palindrome) and add it to the front to create the shortest palindrome string.

The crucial part of the solution is to find the longest palindromic prefix efficiently. We can achieve this by comparing prefixes and suffixes of the string while respecting the palindrome property.

The algorithm uses a hash-based approach to compare the palindromic prefix and suffix, which can be efficiently computed using a rolling hash technique. Here's how it goes:

1. Initialize two hash variables prefix and suffix to 0. These will be used to calculate the hash of the prefix (from the start of the string) and the suffix (from the end of the string) respectively.

2. Use a base for the hash function and a mod to prevent overflow issues with large hash values. Go over each character in the string and update the hash values for both the prefix and suffix.

3. If at any point the hash of the prefix and the hash of the suffix are equal, it indicates that we have a palindrome from the beginning of the string to the current index i.

4. Remember the furthest index idx where the prefix and suffix hashes match (indicating the longest palindromic prefix).

After we identify the longest palindromic prefix, the remaining substring (from idx to the end), when reversed and added to the front of the original string s, will result in the shortest palindrome.

So, the result will be the reverse of the substring from idx to the end concatenated to the original string. If the entire string s is a palindrome, we just return s as it is already the shortest palindrome we can achieve.

Solution Approach

For the implementation of the solution, we leverage a rolling hash polynomial hashing algorithm. The rolling hash technique is an efficient way to compute and compare hash values for substrings quickly. This is particularly useful in our case, where we need to compare prefixes and suffixes of the given string.

Here's a step-by-step explanation of the implementation:

1. We choose a base number, base, for the rolling hash functions and a large prime number, mod, to take the modulus and prevent integer overflow.

2. We initialize prefix and suffix hash values to 0, mul to 1, and idx to 0. The idx will hold the position of the last character of the longest palindrome prefix.

3. We iterate through the string s using a for loop, and on each iteration, we do the following:

   • Update the prefix hash by multiplying the current prefix hash by base and then adding the value of the current character shifted by 1 to avoid 0 in the calculation (ord(c) - ord('a') + 1). We wrap around with modulus mod to manage large numbers.
   • Update the suffix hash by adding to it the value of the current character multiplied by mul (which represents base raised to the power of the character's position). Again, we take the result modulo mod.
   • Update mul by multiplying it by base and taking modulus mod. This step effectively computes base to the power of i modulo mod for the i-th character.
   • If at any point the prefix and suffix hashes are equal, it means we have a palindrome starting from the beginning of the string to the current character. We update idx to i + 1.

4. After the loop completes, if idx is equal to the length of the string, the entire string is a palindrome. We return s in that case.

5. If the whole string isn't a palindrome, we need to create a palindrome by adding characters to the beginning of the string. We do this by taking the substring from idx to the end of s, reversing it, and concatenating it with the original string s.

The final result is generated by the expression s[idx:][::-1] + s, where s[idx:][::-1] creates the needed prefix by reversing the part of the string that is not included in the palindrome and appending it to the front of s to form the shortest palindrome.

This approach is efficient as it has a linear time complexity with respect to the length of the string, and it leverages hashing to quickly identify the longest palindrome prefix.

Example Walkthrough

Let's walk through the solution approach with a small example. Suppose our given string is s = "abacd". We want to find the shortest palindrome by adding characters at the beginning of s.

1. We choose base as a small prime number, for simplicity, let's use 3, and let mod be a large prime number, mod = 10007 to avoid integer overflow.

2. Initialize prefix and suffix hashes to 0, mul to 1, and idx to 0.

3. We'll iterate through the string and compute hashes for the prefix starting from the beginning and the suffix from the end simultaneously:

   i	Character (c)	Prefix Hash (new)	Suffix Hash (new)	mul (new)	idx
   0	'a'	(0*3 + 1) mod 10007	(0 + 1*1) mod 10007	3	1
   1	'b'	(1*3 + 2) mod 10007	(1 + 2*3) mod 10007	3*3	1
   2	'a'	(5*3 + 1) mod 10007	(7 + 1*9) mod 10007	333	3
   3	'c'	(16*3 + 3) mod 10007	(16 + 3*27) mod 10007	333*3	3
   4	'd'	(51*3 + 4) mod 10007	(97 + 4*81) mod 10007	33333	3

   At each step, we are updating the prefix hash by multiplying it by base and adding the current character's value, and updating the suffix hash by adding the current character's value multiplied by mul. mul is updated by multiplying it by base at each step.

   At index i = 2, the prefix and suffix hashes match (16 mod 10007), meaning that we have a palindrome from the start of the string to the character at index 2 ("aba"). So we update idx to 3.

4. After completing the iteration, since idx is not equal to the length of the string, we conclude that the entire string is not a palindrome.

5. To form the shortest palindrome, we take the substring from idx to the end ("acd") and reverse it to get "dca". We then concatenate this reversed substring to the front of the original string s.

The resulting palindrome is "dca" + "abacd" = "dcaabacd", which is the shortest palindrome that we can form by adding characters only at the beginning of the string s. This example illustrates the efficiency of the solution, which finds the longest palindromic prefix and adds the minimum required characters to the front to form the palindrome.

Solution Implementation

Time and Space Complexity

The given Python code implements an algorithm for finding the shortest palindrome by appending characters to the beginning of the given string s. The algorithm is based on calculating hash values from both ends (prefix and suffix) and checking for palindromes.

Time Complexity

The time complexity of this code primarily comes from a single for loop that iterates through each character in the string once. Inside the loop, it computes the prefix and suffix hash values, and compares them to check if they are equal.

Here's the breakdown:

• The hash operations and comparison inside the for loop are O(1) operations as they are done using arithmetic calculations.
• The for loop runs n times, where n is the length of the string s.

Therefore, the time complexity of this code is O(n), where n is the length of the input string.

Space Complexity

The space complexity of the code is determined by the storage used which is independent of the length of the input string s.

Here's the breakdown:

• Variables prefix, suffix, mul, and idx are integers which occupy constant space.
• The slice and reverse operation s[idx:][::-1] creates a new string of at most n-1 characters when the input string is not already a palindrome.

Even though a new string is created in the worst-case scenario, the space complexity is proportional to the input string size which gives us O(n). However, if we consider only the additional space excluding the input and output, the space complexity is actually O(1) since we're only using a fixed amount of additional storage regardless of the input size.

Learn more about how to find time and space complexity quickly using problem constraints.